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Exercise 1566Calculus › Single-variable calculus › Differentiation › Taylor expansionsL2Added 3 Nov 2026

Central binomial coefficients from (1 − 4x)^(−1/2)

#binomial series#generalised binomial coefficients#central binomial

Statement

Show that the coefficient of xnx^n in the expansion of (1−4x)−1/2(1-4x)^{-1/2} at 00 is (2nn)\dbinom{2n}n.

Hint

Use the Taylor coefficient f(n)(0)n!\dfrac{f^{(n)}(0)}{n!} with f(x)=(1−4x)−1/2f(x)=(1-4x)^{-1/2} and compute f(n)f^{(n)} by induction.

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Solution

Derivatives. f′(x)=2(1−4x)−3/2f'(x)=2(1-4x)^{-3/2}, f′′(x)=2⋅6(1−4x)−5/2f''(x)=2\cdot6(1-4x)^{-5/2}, f′′′(x)=2⋅6⋅10(1−4x)−7/2f'''(x)=2\cdot6\cdot10(1-4x)^{-7/2}. By induction,
f(n)(x)=∏k=0n−1(4k+2)  (1−4x)−(2n+1)/2,f^{(n)}(x)=\prod_{k=0}^{n-1}(4k+2)\;(1-4x)^{-(2n+1)/2},
since differentiating (1−4x)−(2n+1)/2(1-4x)^{-(2n+1)/2} multiplies by 2n+12⋅4=4n+2\frac{2n+1}2\cdot4=4n+2.

Coefficient. Therefore
f(n)(0)n!=2⋅6⋅10⋯(4n−2)n!=2n⋅1⋅3⋅5⋯(2n−1)n!.\frac{f^{(n)}(0)}{n!}=\frac{2\cdot6\cdot10\cdots(4n-2)}{n!}=\frac{2^n\cdot1\cdot3\cdot5\cdots(2n-1)}{n!}.
Using 1⋅3⋯(2n−1)=(2n)!2nn!1\cdot3\cdots(2n-1)=\dfrac{(2n)!}{2^nn!}, this equals (2n)!n! n!=(2nn)\dfrac{(2n)!}{n!\,n!}=\dbinom{2n}n.

First terms. (1−4x)−1/2=1+2x+6x2+20x3+70x4+⋯(1-4x)^{-1/2}=1+2x+6x^2+20x^3+70x^4+\cdots ✓.

Remark. This is the generating function of the central binomial coefficients. Applying the same idea to 1−4x\sqrt{1-4x} produces the Catalan numbers.